If $\sin(x) - \cos(x) = 1/3$ then determine $\sin(x)\cos(x)$ 
If
  $$\sin(x) - \cos(x) = \frac{1}{3}$$
  then determine
  $$\sin(x)\cos(x)$$

I know that the expected solution is squaring both sides of equation and solving it this way:
\begin{gather}
\sin^2(x)+\cos^2(x)= 1 \\[4px]
(\sin(x) - \cos(x))^2 = \left(\frac{1}{3}\right)^2 \\[4px]
\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) =\frac{1}{9} \\[4px]
-2\sin(x)\cos(x)=\frac{1}{9} -\sin^2(x)-\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +\sin^2(x)+\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +1\\[4px]
2\sin(x)\cos(x)=\frac{8}{9} \\[4px]
\sin(x)\cos(x)=\frac{4}{9}
\end{gather}
But assume I haven't noticed that I can solve it by squaring both sides in the first place. I can't figure it out how to solve it any other way.
 A: You can use $$\sin x = \frac{2t}{1+t^2}; \cos x = \frac {1-t^2}{1+t^2}$$ 
where $t=\tan (x/2)$, but you just have to know that some value of $t$ works.
Then you get a quadratic to solve for $t$. And once you know $t$ you can solve the problem.
A: You can notice that
$$
\sin x-\cos x=\sqrt{2}\sin(x-\pi/4)
$$
Set $y=x-\pi/4$; then $\sin y=\frac{1}{3\sqrt{2}}$ and
\begin{align}
\sin x\cos x
&=\sin(y+\pi/4)\cos(y+\pi/4) \\[4px]
&=\left(\frac{1}{\sqrt{2}}\cos y+\frac{1}{\sqrt{2}}\sin y\right)
  \left(\frac{1}{\sqrt{2}}\cos y-\frac{1}{\sqrt{2}}\sin y\right) \\[4px]
&=\frac{1}{2}(\cos^2y-\sin^2y) \\[4px]
&=\frac{1}{2}(1-2\sin^2y) \\[4px]
&=\frac{1}{2}\left(1-\frac{2}{(3\sqrt{2})^2}\right)
\end{align}
A: Note that
$$\cos(x + \pi/4) = \cos x \cos(\pi/4) - \sin x \sin(\pi/4) = \frac{1}{\sqrt{2}}(\cos x - \sin x) = -\frac{1}{3\sqrt{2}}.$$
Now, using a double-angle formula:
$$\cos(2x + \pi/2) = 2 \cos^2(x + \pi/4) - 1 = 2 \left( -\frac{1}{3\sqrt{2}}\right)^2 - 1 = -\frac{8}{9}.$$
On the other hand,
$$\cos(2x + \pi/2) = \cos(2x) \cos(\pi/2) - \sin(2x) \sin(\pi/2) = -\sin(2x) = -2 \sin x \cos x.$$
Equating the two, $-2 \sin x \cos x = -\frac{8}{9}$.

In general, this method of using a sum-of-angles formula in reverse is very often useful in analyzing expressions of the form $A \cos x + B \sin x$ when $A$ and $B$ are constants, and is a nice tool to keep available in your toolbox.
A: To save typing, let $s = \sin x$ and $c = \cos x$.  Then we have two equations in two variables:
$$
\begin{align}
s^2 + c^2 &= 1\\
s - c &= \frac13.
\end{align}
$$
A widely applicable approach in such a situation is to solve one equation for one of the variables, and use that result to substitute into the other.  Here, the second equation can easily be solved for $s$, yielding $s = c + \frac13$.  Substituting for $s$ in the first equation gives a quadratic equation in $c$:
$$
2c^2 + \frac23 c - \frac89 = 0.
$$
The roots of this quadratic are
$$
c = \frac{-1+\sqrt{17}}{6}, \frac{-1-\sqrt{17}}{6}.
$$
Since $s = c + \frac13$, the corresponding values for $s$ are
$$
s = \frac{1+\sqrt{17}}{6}, \frac{1-\sqrt{17}}{6}.
$$
Multiplying either pair of corresponding values gives the result $sc = \frac49$.
This method is much more laborious than the original solution, but requires no cleverness or flash of insight.
